AoPS Academy
satisfies
for some 
and for all 
, then there exists a unique solution 
of the functional equation 
for all such that
for all 
.
+1 w
be distinct polynomials of degree 
with non-zero coefficients. Suppose that they have 
common real roots counting multiplicity and 
common coefficients. Determine the maximum possible value of 
.
such that:
is strictly increasing,
such that 
for all 
,
, there exists 
such that![\[
f(y) = \frac{f(x) + f(x + 2024)}{2}.
\]](http://latex.artofproblemsolving.com/b/e/2/be26213154bb74bd5a35b8d160011351871bfa9b.png)
Proposed by Pavle Martinović
, there is a 
so that![\[2^n + 3^n + 6^n - 1 \equiv 0\pmod{p}\]](http://latex.artofproblemsolving.com/6/a/d/6ad29e27ec0af4da7cf507a4e6dd028d3b82747b.png)
Let 
.![\[\frac{1}{2} + \frac{1}{3} + \frac{1}{6} - 1 \equiv 0\pmod{p}\]](http://latex.artofproblemsolving.com/6/b/a/6ba086076d7eeff9592a9116904281db53aa0def.png)
so 
is divisible by every prime at some point.
.
contains all primes.. If 
then we have 
which is true for any . If 
then we have 
which is true for all even . So from now on we consider only such that 
. divides some then we have 
. Note that since , we can multiply both sides by the inverse of 
to get 
. Since is coprime to 
, there exists an order for 
, thus we may take to be that order and we are done.
is the only positive integer relatively prime to all terms of . Since this is obviously true, consider some positive integer 
. If 
is a power of 
or 
, then we can take 
to get 
which is not relatively prime to . Otherwise, consider a prime 
that divides . Taking 
, we have
so 
, implying that there always exists at least term of that shares a prime factor with every positive integer greater than , as claimed. 
, which clearly works. We will prove that there are no primes relatively prime to all . There are two cases.
, : Here, we can take 
.
, : Here, we will prove that 
Note that 
which completes the problem.
which obviously works. It clearly suffices to show that 
. The conclusion for is true for all and for it is true for all even . For other , consider . So, 
, so we are done.
and . We have that 
modulo so all primes greater than divide atleast one term of this sequence so we are done. obviously divides all terms and divides all those with even.
so multiples of 
don't work. Now, suppose that is a prime. Then by FLT 
Therefore is the only solution.
so multiples of and do not satisfy the desired condition. Further, for all primes , by Fermat's Little Theorem we have,![\[2^{p-2}+3^{p-2}+6^{p-2}-1\equiv \frac{1}{2}+\frac{1}{3}+\frac{1}{6}-1 \equiv 0 \pmod{p}\]](http://latex.artofproblemsolving.com/6/7/2/672f7522379f8e28a4f5f8d901177aa72acc5282.png)
Thus, 
for all primes . Thus, any integer with a prime factor greater than also violates the condition, implying that the only solution is .![\[ a_n=2^n+3^n+6^n -1,\ n\geq 1. \]](http://latex.artofproblemsolving.com/6/4/9/6499d8f992f3e72765cab1f3afe15df99d01b12a.png)
for which 
satisfy this condition, but the cases where 
are messier. In particular any positive integer of the form 
does not satisfy this condition.
Prove that
is divisible by 
,
)
and so it is divisible by 6.
because it doesn't seem ok (I mean 
would mean that 5 divides 15/2 from the definition of congruence).. So I wrote it like this 
and then we get 
which is obviously divisible by p (for p>=5), and beacuse 
and 
and 
such that 
modulo 
,
.
! I think this is the easiest Number theory problem in IMO .
: 
is divisible by 
(It's proved easily by Fermat Little Theorem)
,
are 
and 
.
!
,
is a complex number so we are done
![\[2^{p-2}+3^{p-2}+6^{p-2}\equiv2^{-1}+3^{-1}+6^{-1}\equiv1\pmod{p}.\]](http://latex.artofproblemsolving.com/d/e/a/deabeef3910647848a5e1104e94d4999f44bcc38.png)
for all 
,![\[a_{p-2} \equiv \frac{1}{2}+\frac{1}{3}+\frac{1}{6}-1 \equiv 0\pmod{p}\]](http://latex.artofproblemsolving.com/5/2/3/523533c71a0b86d911bd502c4e4f364aeb4cb041.png)
for all 
.
and 
respectively. Thus, the only such 
,
,
,
,
and 
By FLT, we can say 
where 
Because 
we can say 
so we are done.